If x and y are positive integers such that y is a multiple of 5…

If \(x\) and \(y\) are positive integers such that \(y\) is a multiple of 5 and \(3x + 4y = 200\), then \(x\) must be a multiple of which of the following?

(A) 3

(B) 6

(C) 7

(D) 8

(E) 10

An obvious strategy for most students is to start substituting multiples of 5 for \(y\). This wouldn’t be my gut reaction, but it works. I think this works best if we isolate \(x\) first:

\(3x + 4y = 200\)

Subtract \(4y\) from both sides:

\(3x = 200 – 4y\)

Divide both sides by 3:

\(x = \frac{200 – 4y}{3}\)

If \(y = 5\), we get \(x = \frac{200 – 4(5)}{3} = \frac{200 – 20}{3} = \frac{180}{3} = 60\), and \(x\) is a multiple of 3, 6, and 10, but not 7 or 8, so we can eliminate C and D.

It’s easy to see that if \(y\) is 10 or 15, \(200 – 4y\) is not divisible by 3. This is the primary benefit of isolating \(x\) before checking multiples of \(y\).

If \(y = 20\), we get \(x = \frac{200 – 4(20)}{3} = \frac{200 – 80}{3} = \frac{120}{3} = 40\), and \(x\) is a multiple of 10, but not 3 or 6, so we can eliminate A and D.

The correct answer is E.

My gut reaction would be to substitute \(5n\) for \(y\). After some algebra, we get

\(3x = 200 – 4(5n) = 200 – 20n\)

Factor 20 out of the right-hand-side to get

\(3x = 20(10 – n)\)

Since we’re dealing exclusively with integers and 3 is not a divisor of 20, the prime factors of 20 must come from the \(x\). This is a bit more complex than it sounds at first, so take a sec to think about it…. Got it? Good! So, \(x\) must be divisible by 20, which means it must be divisible by 10, but not necessarily 3, 6, 7, or 8.